1.1 --- /dev/null Thu Jan 01 00:00:00 1970 +0000
1.2 +++ b/src/glpnpp04.c Mon Dec 06 13:09:21 2010 +0100
1.3 @@ -0,0 +1,1414 @@
1.4 +/* glpnpp04.c */
1.5 +
1.6 +/***********************************************************************
1.7 +* This code is part of GLPK (GNU Linear Programming Kit).
1.8 +*
1.9 +* Copyright (C) 2000, 2001, 2002, 2003, 2004, 2005, 2006, 2007, 2008,
1.10 +* 2009, 2010 Andrew Makhorin, Department for Applied Informatics,
1.11 +* Moscow Aviation Institute, Moscow, Russia. All rights reserved.
1.12 +* E-mail: <mao@gnu.org>.
1.13 +*
1.14 +* GLPK is free software: you can redistribute it and/or modify it
1.15 +* under the terms of the GNU General Public License as published by
1.16 +* the Free Software Foundation, either version 3 of the License, or
1.17 +* (at your option) any later version.
1.18 +*
1.19 +* GLPK is distributed in the hope that it will be useful, but WITHOUT
1.20 +* ANY WARRANTY; without even the implied warranty of MERCHANTABILITY
1.21 +* or FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public
1.22 +* License for more details.
1.23 +*
1.24 +* You should have received a copy of the GNU General Public License
1.25 +* along with GLPK. If not, see <http://www.gnu.org/licenses/>.
1.26 +***********************************************************************/
1.27 +
1.28 +#include "glpnpp.h"
1.29 +
1.30 +/***********************************************************************
1.31 +* NAME
1.32 +*
1.33 +* npp_binarize_prob - binarize MIP problem
1.34 +*
1.35 +* SYNOPSIS
1.36 +*
1.37 +* #include "glpnpp.h"
1.38 +* int npp_binarize_prob(NPP *npp);
1.39 +*
1.40 +* DESCRIPTION
1.41 +*
1.42 +* The routine npp_binarize_prob replaces in the original MIP problem
1.43 +* every integer variable:
1.44 +*
1.45 +* l[q] <= x[q] <= u[q], (1)
1.46 +*
1.47 +* where l[q] < u[q], by an equivalent sum of binary variables.
1.48 +*
1.49 +* RETURNS
1.50 +*
1.51 +* The routine returns the number of integer variables for which the
1.52 +* transformation failed, because u[q] - l[q] > d_max.
1.53 +*
1.54 +* PROBLEM TRANSFORMATION
1.55 +*
1.56 +* If variable x[q] has non-zero lower bound, it is first processed
1.57 +* with the routine npp_lbnd_col. Thus, we can assume that:
1.58 +*
1.59 +* 0 <= x[q] <= u[q]. (2)
1.60 +*
1.61 +* If u[q] = 1, variable x[q] is already binary, so further processing
1.62 +* is not needed. Let, therefore, that 2 <= u[q] <= d_max, and n be a
1.63 +* smallest integer such that u[q] <= 2^n - 1 (n >= 2, since u[q] >= 2).
1.64 +* Then variable x[q] can be replaced by the following sum:
1.65 +*
1.66 +* n-1
1.67 +* x[q] = sum 2^k x[k], (3)
1.68 +* k=0
1.69 +*
1.70 +* where x[k] are binary columns (variables). If u[q] < 2^n - 1, the
1.71 +* following additional inequality constraint must be also included in
1.72 +* the transformed problem:
1.73 +*
1.74 +* n-1
1.75 +* sum 2^k x[k] <= u[q]. (4)
1.76 +* k=0
1.77 +*
1.78 +* Note: Assuming that in the transformed problem x[q] becomes binary
1.79 +* variable x[0], this transformation causes new n-1 binary variables
1.80 +* to appear.
1.81 +*
1.82 +* Substituting x[q] from (3) to the objective row gives:
1.83 +*
1.84 +* z = sum c[j] x[j] + c[0] =
1.85 +* j
1.86 +*
1.87 +* = sum c[j] x[j] + c[q] x[q] + c[0] =
1.88 +* j!=q
1.89 +* n-1
1.90 +* = sum c[j] x[j] + c[q] sum 2^k x[k] + c[0] =
1.91 +* j!=q k=0
1.92 +* n-1
1.93 +* = sum c[j] x[j] + sum c[k] x[k] + c[0],
1.94 +* j!=q k=0
1.95 +*
1.96 +* where:
1.97 +*
1.98 +* c[k] = 2^k c[q], k = 0, ..., n-1. (5)
1.99 +*
1.100 +* And substituting x[q] from (3) to i-th constraint row i gives:
1.101 +*
1.102 +* L[i] <= sum a[i,j] x[j] <= U[i] ==>
1.103 +* j
1.104 +*
1.105 +* L[i] <= sum a[i,j] x[j] + a[i,q] x[q] <= U[i] ==>
1.106 +* j!=q
1.107 +* n-1
1.108 +* L[i] <= sum a[i,j] x[j] + a[i,q] sum 2^k x[k] <= U[i] ==>
1.109 +* j!=q k=0
1.110 +* n-1
1.111 +* L[i] <= sum a[i,j] x[j] + sum a[i,k] x[k] <= U[i],
1.112 +* j!=q k=0
1.113 +*
1.114 +* where:
1.115 +*
1.116 +* a[i,k] = 2^k a[i,q], k = 0, ..., n-1. (6)
1.117 +*
1.118 +* RECOVERING SOLUTION
1.119 +*
1.120 +* Value of variable x[q] is computed with formula (3). */
1.121 +
1.122 +struct binarize
1.123 +{ int q;
1.124 + /* column reference number for x[q] = x[0] */
1.125 + int j;
1.126 + /* column reference number for x[1]; x[2] has reference number
1.127 + j+1, x[3] - j+2, etc. */
1.128 + int n;
1.129 + /* total number of binary variables, n >= 2 */
1.130 +};
1.131 +
1.132 +static int rcv_binarize_prob(NPP *npp, void *info);
1.133 +
1.134 +int npp_binarize_prob(NPP *npp)
1.135 +{ /* binarize MIP problem */
1.136 + struct binarize *info;
1.137 + NPPROW *row;
1.138 + NPPCOL *col, *bin;
1.139 + NPPAIJ *aij;
1.140 + int u, n, k, temp, nfails, nvars, nbins, nrows;
1.141 + /* new variables will be added to the end of the column list, so
1.142 + we go from the end to beginning of the column list */
1.143 + nfails = nvars = nbins = nrows = 0;
1.144 + for (col = npp->c_tail; col != NULL; col = col->prev)
1.145 + { /* skip continuous variable */
1.146 + if (!col->is_int) continue;
1.147 + /* skip fixed variable */
1.148 + if (col->lb == col->ub) continue;
1.149 + /* skip binary variable */
1.150 + if (col->lb == 0.0 && col->ub == 1.0) continue;
1.151 + /* check if the transformation is applicable */
1.152 + if (col->lb < -1e6 || col->ub > +1e6 ||
1.153 + col->ub - col->lb > 4095.0)
1.154 + { /* unfortunately, not */
1.155 + nfails++;
1.156 + continue;
1.157 + }
1.158 + /* process integer non-binary variable x[q] */
1.159 + nvars++;
1.160 + /* make x[q] non-negative, if its lower bound is non-zero */
1.161 + if (col->lb != 0.0)
1.162 + npp_lbnd_col(npp, col);
1.163 + /* now 0 <= x[q] <= u[q] */
1.164 + xassert(col->lb == 0.0);
1.165 + u = (int)col->ub;
1.166 + xassert(col->ub == (double)u);
1.167 + /* if x[q] is binary, further processing is not needed */
1.168 + if (u == 1) continue;
1.169 + /* determine smallest n such that u <= 2^n - 1 (thus, n is the
1.170 + number of binary variables needed) */
1.171 + n = 2, temp = 4;
1.172 + while (u >= temp)
1.173 + n++, temp += temp;
1.174 + nbins += n;
1.175 + /* create transformation stack entry */
1.176 + info = npp_push_tse(npp,
1.177 + rcv_binarize_prob, sizeof(struct binarize));
1.178 + info->q = col->j;
1.179 + info->j = 0; /* will be set below */
1.180 + info->n = n;
1.181 + /* if u < 2^n - 1, we need one additional row for (4) */
1.182 + if (u < temp - 1)
1.183 + { row = npp_add_row(npp), nrows++;
1.184 + row->lb = -DBL_MAX, row->ub = u;
1.185 + }
1.186 + else
1.187 + row = NULL;
1.188 + /* in the transformed problem variable x[q] becomes binary
1.189 + variable x[0], so its objective and constraint coefficients
1.190 + are not changed */
1.191 + col->ub = 1.0;
1.192 + /* include x[0] into constraint (4) */
1.193 + if (row != NULL)
1.194 + npp_add_aij(npp, row, col, 1.0);
1.195 + /* add other binary variables x[1], ..., x[n-1] */
1.196 + for (k = 1, temp = 2; k < n; k++, temp += temp)
1.197 + { /* add new binary variable x[k] */
1.198 + bin = npp_add_col(npp);
1.199 + bin->is_int = 1;
1.200 + bin->lb = 0.0, bin->ub = 1.0;
1.201 + bin->coef = (double)temp * col->coef;
1.202 + /* store column reference number for x[1] */
1.203 + if (info->j == 0)
1.204 + info->j = bin->j;
1.205 + else
1.206 + xassert(info->j + (k-1) == bin->j);
1.207 + /* duplicate constraint coefficients for x[k]; this also
1.208 + automatically includes x[k] into constraint (4) */
1.209 + for (aij = col->ptr; aij != NULL; aij = aij->c_next)
1.210 + npp_add_aij(npp, aij->row, bin, (double)temp * aij->val);
1.211 + }
1.212 + }
1.213 + if (nvars > 0)
1.214 + xprintf("%d integer variable(s) were replaced by %d binary one"
1.215 + "s\n", nvars, nbins);
1.216 + if (nrows > 0)
1.217 + xprintf("%d row(s) were added due to binarization\n", nrows);
1.218 + if (nfails > 0)
1.219 + xprintf("Binarization failed for %d integer variable(s)\n",
1.220 + nfails);
1.221 + return nfails;
1.222 +}
1.223 +
1.224 +static int rcv_binarize_prob(NPP *npp, void *_info)
1.225 +{ /* recovery binarized variable */
1.226 + struct binarize *info = _info;
1.227 + int k, temp;
1.228 + double sum;
1.229 + /* compute value of x[q]; see formula (3) */
1.230 + sum = npp->c_value[info->q];
1.231 + for (k = 1, temp = 2; k < info->n; k++, temp += temp)
1.232 + sum += (double)temp * npp->c_value[info->j + (k-1)];
1.233 + npp->c_value[info->q] = sum;
1.234 + return 0;
1.235 +}
1.236 +
1.237 +/**********************************************************************/
1.238 +
1.239 +struct elem
1.240 +{ /* linear form element a[j] x[j] */
1.241 + double aj;
1.242 + /* non-zero coefficient value */
1.243 + NPPCOL *xj;
1.244 + /* pointer to variable (column) */
1.245 + struct elem *next;
1.246 + /* pointer to another term */
1.247 +};
1.248 +
1.249 +static struct elem *copy_form(NPP *npp, NPPROW *row, double s)
1.250 +{ /* copy linear form */
1.251 + NPPAIJ *aij;
1.252 + struct elem *ptr, *e;
1.253 + ptr = NULL;
1.254 + for (aij = row->ptr; aij != NULL; aij = aij->r_next)
1.255 + { e = dmp_get_atom(npp->pool, sizeof(struct elem));
1.256 + e->aj = s * aij->val;
1.257 + e->xj = aij->col;
1.258 + e->next = ptr;
1.259 + ptr = e;
1.260 + }
1.261 + return ptr;
1.262 +}
1.263 +
1.264 +static void drop_form(NPP *npp, struct elem *ptr)
1.265 +{ /* drop linear form */
1.266 + struct elem *e;
1.267 + while (ptr != NULL)
1.268 + { e = ptr;
1.269 + ptr = e->next;
1.270 + dmp_free_atom(npp->pool, e, sizeof(struct elem));
1.271 + }
1.272 + return;
1.273 +}
1.274 +
1.275 +/***********************************************************************
1.276 +* NAME
1.277 +*
1.278 +* npp_is_packing - test if constraint is packing inequality
1.279 +*
1.280 +* SYNOPSIS
1.281 +*
1.282 +* #include "glpnpp.h"
1.283 +* int npp_is_packing(NPP *npp, NPPROW *row);
1.284 +*
1.285 +* RETURNS
1.286 +*
1.287 +* If the specified row (constraint) is packing inequality (see below),
1.288 +* the routine npp_is_packing returns non-zero. Otherwise, it returns
1.289 +* zero.
1.290 +*
1.291 +* PACKING INEQUALITIES
1.292 +*
1.293 +* In canonical format the packing inequality is the following:
1.294 +*
1.295 +* sum x[j] <= 1, (1)
1.296 +* j in J
1.297 +*
1.298 +* where all variables x[j] are binary. This inequality expresses the
1.299 +* condition that in any integer feasible solution at most one variable
1.300 +* from set J can take non-zero (unity) value while other variables
1.301 +* must be equal to zero. W.l.o.g. it is assumed that |J| >= 2, because
1.302 +* if J is empty or |J| = 1, the inequality (1) is redundant.
1.303 +*
1.304 +* In general case the packing inequality may include original variables
1.305 +* x[j] as well as their complements x~[j]:
1.306 +*
1.307 +* sum x[j] + sum x~[j] <= 1, (2)
1.308 +* j in Jp j in Jn
1.309 +*
1.310 +* where Jp and Jn are not intersected. Therefore, using substitution
1.311 +* x~[j] = 1 - x[j] gives the packing inequality in generalized format:
1.312 +*
1.313 +* sum x[j] - sum x[j] <= 1 - |Jn|. (3)
1.314 +* j in Jp j in Jn */
1.315 +
1.316 +int npp_is_packing(NPP *npp, NPPROW *row)
1.317 +{ /* test if constraint is packing inequality */
1.318 + NPPCOL *col;
1.319 + NPPAIJ *aij;
1.320 + int b;
1.321 + xassert(npp == npp);
1.322 + if (!(row->lb == -DBL_MAX && row->ub != +DBL_MAX))
1.323 + return 0;
1.324 + b = 1;
1.325 + for (aij = row->ptr; aij != NULL; aij = aij->r_next)
1.326 + { col = aij->col;
1.327 + if (!(col->is_int && col->lb == 0.0 && col->ub == 1.0))
1.328 + return 0;
1.329 + if (aij->val == +1.0)
1.330 + ;
1.331 + else if (aij->val == -1.0)
1.332 + b--;
1.333 + else
1.334 + return 0;
1.335 + }
1.336 + if (row->ub != (double)b) return 0;
1.337 + return 1;
1.338 +}
1.339 +
1.340 +/***********************************************************************
1.341 +* NAME
1.342 +*
1.343 +* npp_hidden_packing - identify hidden packing inequality
1.344 +*
1.345 +* SYNOPSIS
1.346 +*
1.347 +* #include "glpnpp.h"
1.348 +* int npp_hidden_packing(NPP *npp, NPPROW *row);
1.349 +*
1.350 +* DESCRIPTION
1.351 +*
1.352 +* The routine npp_hidden_packing processes specified inequality
1.353 +* constraint, which includes only binary variables, and the number of
1.354 +* the variables is not less than two. If the original inequality is
1.355 +* equivalent to a packing inequality, the routine replaces it by this
1.356 +* equivalent inequality. If the original constraint is double-sided
1.357 +* inequality, it is replaced by a pair of single-sided inequalities,
1.358 +* if necessary.
1.359 +*
1.360 +* RETURNS
1.361 +*
1.362 +* If the original inequality constraint was replaced by equivalent
1.363 +* packing inequality, the routine npp_hidden_packing returns non-zero.
1.364 +* Otherwise, it returns zero.
1.365 +*
1.366 +* PROBLEM TRANSFORMATION
1.367 +*
1.368 +* Consider an inequality constraint:
1.369 +*
1.370 +* sum a[j] x[j] <= b, (1)
1.371 +* j in J
1.372 +*
1.373 +* where all variables x[j] are binary, and |J| >= 2. (In case of '>='
1.374 +* inequality it can be transformed to '<=' format by multiplying both
1.375 +* its sides by -1.)
1.376 +*
1.377 +* Let Jp = {j: a[j] > 0}, Jn = {j: a[j] < 0}. Performing substitution
1.378 +* x[j] = 1 - x~[j] for all j in Jn, we have:
1.379 +*
1.380 +* sum a[j] x[j] <= b ==>
1.381 +* j in J
1.382 +*
1.383 +* sum a[j] x[j] + sum a[j] x[j] <= b ==>
1.384 +* j in Jp j in Jn
1.385 +*
1.386 +* sum a[j] x[j] + sum a[j] (1 - x~[j]) <= b ==>
1.387 +* j in Jp j in Jn
1.388 +*
1.389 +* sum a[j] x[j] - sum a[j] x~[j] <= b - sum a[j].
1.390 +* j in Jp j in Jn j in Jn
1.391 +*
1.392 +* Thus, meaning the transformation above, we can assume that in
1.393 +* inequality (1) all coefficients a[j] are positive. Moreover, we can
1.394 +* assume that a[j] <= b. In fact, let a[j] > b; then the following
1.395 +* three cases are possible:
1.396 +*
1.397 +* 1) b < 0. In this case inequality (1) is infeasible, so the problem
1.398 +* has no feasible solution (see the routine npp_analyze_row);
1.399 +*
1.400 +* 2) b = 0. In this case inequality (1) is a forcing inequality on its
1.401 +* upper bound (see the routine npp_forcing row), from which it
1.402 +* follows that all variables x[j] should be fixed at zero;
1.403 +*
1.404 +* 3) b > 0. In this case inequality (1) defines an implied zero upper
1.405 +* bound for variable x[j] (see the routine npp_implied_bounds), from
1.406 +* which it follows that x[j] should be fixed at zero.
1.407 +*
1.408 +* It is assumed that all three cases listed above have been recognized
1.409 +* by the routine npp_process_prob, which performs basic MIP processing
1.410 +* prior to a call the routine npp_hidden_packing. So, if one of these
1.411 +* cases occurs, we should just skip processing such constraint.
1.412 +*
1.413 +* Thus, let 0 < a[j] <= b. Then it is obvious that constraint (1) is
1.414 +* equivalent to packing inquality only if:
1.415 +*
1.416 +* a[j] + a[k] > b + eps (2)
1.417 +*
1.418 +* for all j, k in J, j != k, where eps is an absolute tolerance for
1.419 +* row (linear form) value. Checking the condition (2) for all j and k,
1.420 +* j != k, requires time O(|J|^2). However, this time can be reduced to
1.421 +* O(|J|), if use minimal a[j] and a[k], in which case it is sufficient
1.422 +* to check the condition (2) only once.
1.423 +*
1.424 +* Once the original inequality (1) is replaced by equivalent packing
1.425 +* inequality, we need to perform back substitution x~[j] = 1 - x[j] for
1.426 +* all j in Jn (see above).
1.427 +*
1.428 +* RECOVERING SOLUTION
1.429 +*
1.430 +* None needed. */
1.431 +
1.432 +static int hidden_packing(NPP *npp, struct elem *ptr, double *_b)
1.433 +{ /* process inequality constraint: sum a[j] x[j] <= b;
1.434 + 0 - specified row is NOT hidden packing inequality;
1.435 + 1 - specified row is packing inequality;
1.436 + 2 - specified row is hidden packing inequality. */
1.437 + struct elem *e, *ej, *ek;
1.438 + int neg;
1.439 + double b = *_b, eps;
1.440 + xassert(npp == npp);
1.441 + /* a[j] must be non-zero, x[j] must be binary, for all j in J */
1.442 + for (e = ptr; e != NULL; e = e->next)
1.443 + { xassert(e->aj != 0.0);
1.444 + xassert(e->xj->is_int);
1.445 + xassert(e->xj->lb == 0.0 && e->xj->ub == 1.0);
1.446 + }
1.447 + /* check if the specified inequality constraint already has the
1.448 + form of packing inequality */
1.449 + neg = 0; /* neg is |Jn| */
1.450 + for (e = ptr; e != NULL; e = e->next)
1.451 + { if (e->aj == +1.0)
1.452 + ;
1.453 + else if (e->aj == -1.0)
1.454 + neg++;
1.455 + else
1.456 + break;
1.457 + }
1.458 + if (e == NULL)
1.459 + { /* all coefficients a[j] are +1 or -1; check rhs b */
1.460 + if (b == (double)(1 - neg))
1.461 + { /* it is packing inequality; no processing is needed */
1.462 + return 1;
1.463 + }
1.464 + }
1.465 + /* substitute x[j] = 1 - x~[j] for all j in Jn to make all a[j]
1.466 + positive; the result is a~[j] = |a[j]| and new rhs b */
1.467 + for (e = ptr; e != NULL; e = e->next)
1.468 + if (e->aj < 0) b -= e->aj;
1.469 + /* now a[j] > 0 for all j in J (actually |a[j]| are used) */
1.470 + /* if a[j] > b, skip processing--this case must not appear */
1.471 + for (e = ptr; e != NULL; e = e->next)
1.472 + if (fabs(e->aj) > b) return 0;
1.473 + /* now 0 < a[j] <= b for all j in J */
1.474 + /* find two minimal coefficients a[j] and a[k], j != k */
1.475 + ej = NULL;
1.476 + for (e = ptr; e != NULL; e = e->next)
1.477 + if (ej == NULL || fabs(ej->aj) > fabs(e->aj)) ej = e;
1.478 + xassert(ej != NULL);
1.479 + ek = NULL;
1.480 + for (e = ptr; e != NULL; e = e->next)
1.481 + if (e != ej)
1.482 + if (ek == NULL || fabs(ek->aj) > fabs(e->aj)) ek = e;
1.483 + xassert(ek != NULL);
1.484 + /* the specified constraint is equivalent to packing inequality
1.485 + iff a[j] + a[k] > b + eps */
1.486 + eps = 1e-3 + 1e-6 * fabs(b);
1.487 + if (fabs(ej->aj) + fabs(ek->aj) <= b + eps) return 0;
1.488 + /* perform back substitution x~[j] = 1 - x[j] and construct the
1.489 + final equivalent packing inequality in generalized format */
1.490 + b = 1.0;
1.491 + for (e = ptr; e != NULL; e = e->next)
1.492 + { if (e->aj > 0.0)
1.493 + e->aj = +1.0;
1.494 + else /* e->aj < 0.0 */
1.495 + e->aj = -1.0, b -= 1.0;
1.496 + }
1.497 + *_b = b;
1.498 + return 2;
1.499 +}
1.500 +
1.501 +int npp_hidden_packing(NPP *npp, NPPROW *row)
1.502 +{ /* identify hidden packing inequality */
1.503 + NPPROW *copy;
1.504 + NPPAIJ *aij;
1.505 + struct elem *ptr, *e;
1.506 + int kase, ret, count = 0;
1.507 + double b;
1.508 + /* the row must be inequality constraint */
1.509 + xassert(row->lb < row->ub);
1.510 + for (kase = 0; kase <= 1; kase++)
1.511 + { if (kase == 0)
1.512 + { /* process row upper bound */
1.513 + if (row->ub == +DBL_MAX) continue;
1.514 + ptr = copy_form(npp, row, +1.0);
1.515 + b = + row->ub;
1.516 + }
1.517 + else
1.518 + { /* process row lower bound */
1.519 + if (row->lb == -DBL_MAX) continue;
1.520 + ptr = copy_form(npp, row, -1.0);
1.521 + b = - row->lb;
1.522 + }
1.523 + /* now the inequality has the form "sum a[j] x[j] <= b" */
1.524 + ret = hidden_packing(npp, ptr, &b);
1.525 + xassert(0 <= ret && ret <= 2);
1.526 + if (kase == 1 && ret == 1 || ret == 2)
1.527 + { /* the original inequality has been identified as hidden
1.528 + packing inequality */
1.529 + count++;
1.530 +#ifdef GLP_DEBUG
1.531 + xprintf("Original constraint:\n");
1.532 + for (aij = row->ptr; aij != NULL; aij = aij->r_next)
1.533 + xprintf(" %+g x%d", aij->val, aij->col->j);
1.534 + if (row->lb != -DBL_MAX) xprintf(", >= %g", row->lb);
1.535 + if (row->ub != +DBL_MAX) xprintf(", <= %g", row->ub);
1.536 + xprintf("\n");
1.537 + xprintf("Equivalent packing inequality:\n");
1.538 + for (e = ptr; e != NULL; e = e->next)
1.539 + xprintf(" %sx%d", e->aj > 0.0 ? "+" : "-", e->xj->j);
1.540 + xprintf(", <= %g\n", b);
1.541 +#endif
1.542 + if (row->lb == -DBL_MAX || row->ub == +DBL_MAX)
1.543 + { /* the original row is single-sided inequality; no copy
1.544 + is needed */
1.545 + copy = NULL;
1.546 + }
1.547 + else
1.548 + { /* the original row is double-sided inequality; we need
1.549 + to create its copy for other bound before replacing it
1.550 + with the equivalent inequality */
1.551 + copy = npp_add_row(npp);
1.552 + if (kase == 0)
1.553 + { /* the copy is for lower bound */
1.554 + copy->lb = row->lb, copy->ub = +DBL_MAX;
1.555 + }
1.556 + else
1.557 + { /* the copy is for upper bound */
1.558 + copy->lb = -DBL_MAX, copy->ub = row->ub;
1.559 + }
1.560 + /* copy original row coefficients */
1.561 + for (aij = row->ptr; aij != NULL; aij = aij->r_next)
1.562 + npp_add_aij(npp, copy, aij->col, aij->val);
1.563 + }
1.564 + /* replace the original inequality by equivalent one */
1.565 + npp_erase_row(npp, row);
1.566 + row->lb = -DBL_MAX, row->ub = b;
1.567 + for (e = ptr; e != NULL; e = e->next)
1.568 + npp_add_aij(npp, row, e->xj, e->aj);
1.569 + /* continue processing lower bound for the copy */
1.570 + if (copy != NULL) row = copy;
1.571 + }
1.572 + drop_form(npp, ptr);
1.573 + }
1.574 + return count;
1.575 +}
1.576 +
1.577 +/***********************************************************************
1.578 +* NAME
1.579 +*
1.580 +* npp_implied_packing - identify implied packing inequality
1.581 +*
1.582 +* SYNOPSIS
1.583 +*
1.584 +* #include "glpnpp.h"
1.585 +* int npp_implied_packing(NPP *npp, NPPROW *row, int which,
1.586 +* NPPCOL *var[], char set[]);
1.587 +*
1.588 +* DESCRIPTION
1.589 +*
1.590 +* The routine npp_implied_packing processes specified row (constraint)
1.591 +* of general format:
1.592 +*
1.593 +* L <= sum a[j] x[j] <= U. (1)
1.594 +* j
1.595 +*
1.596 +* If which = 0, only lower bound L, which must exist, is considered,
1.597 +* while upper bound U is ignored. Similarly, if which = 1, only upper
1.598 +* bound U, which must exist, is considered, while lower bound L is
1.599 +* ignored. Thus, if the specified row is a double-sided inequality or
1.600 +* equality constraint, this routine should be called twice for both
1.601 +* lower and upper bounds.
1.602 +*
1.603 +* The routine npp_implied_packing attempts to find a non-trivial (i.e.
1.604 +* having not less than two binary variables) packing inequality:
1.605 +*
1.606 +* sum x[j] - sum x[j] <= 1 - |Jn|, (2)
1.607 +* j in Jp j in Jn
1.608 +*
1.609 +* which is relaxation of the constraint (1) in the sense that any
1.610 +* solution satisfying to that constraint also satisfies to the packing
1.611 +* inequality (2). If such relaxation exists, the routine stores
1.612 +* pointers to descriptors of corresponding binary variables and their
1.613 +* flags, resp., to locations var[1], var[2], ..., var[len] and set[1],
1.614 +* set[2], ..., set[len], where set[j] = 0 means that j in Jp and
1.615 +* set[j] = 1 means that j in Jn.
1.616 +*
1.617 +* RETURNS
1.618 +*
1.619 +* The routine npp_implied_packing returns len, which is the total
1.620 +* number of binary variables in the packing inequality found, len >= 2.
1.621 +* However, if the relaxation does not exist, the routine returns zero.
1.622 +*
1.623 +* ALGORITHM
1.624 +*
1.625 +* If which = 0, the constraint coefficients (1) are multiplied by -1
1.626 +* and b is assigned -L; if which = 1, the constraint coefficients (1)
1.627 +* are not changed and b is assigned +U. In both cases the specified
1.628 +* constraint gets the following format:
1.629 +*
1.630 +* sum a[j] x[j] <= b. (3)
1.631 +* j
1.632 +*
1.633 +* (Note that (3) is a relaxation of (1), because one of bounds L or U
1.634 +* is ignored.)
1.635 +*
1.636 +* Let J be set of binary variables, Kp be set of non-binary (integer
1.637 +* or continuous) variables with a[j] > 0, and Kn be set of non-binary
1.638 +* variables with a[j] < 0. Then the inequality (3) can be written as
1.639 +* follows:
1.640 +*
1.641 +* sum a[j] x[j] <= b - sum a[j] x[j] - sum a[j] x[j]. (4)
1.642 +* j in J j in Kp j in Kn
1.643 +*
1.644 +* To get rid of non-binary variables we can replace the inequality (4)
1.645 +* by the following relaxed inequality:
1.646 +*
1.647 +* sum a[j] x[j] <= b~, (5)
1.648 +* j in J
1.649 +*
1.650 +* where:
1.651 +*
1.652 +* b~ = sup(b - sum a[j] x[j] - sum a[j] x[j]) =
1.653 +* j in Kp j in Kn
1.654 +*
1.655 +* = b - inf sum a[j] x[j] - inf sum a[j] x[j] = (6)
1.656 +* j in Kp j in Kn
1.657 +*
1.658 +* = b - sum a[j] l[j] - sum a[j] u[j].
1.659 +* j in Kp j in Kn
1.660 +*
1.661 +* Note that if lower bound l[j] (if j in Kp) or upper bound u[j]
1.662 +* (if j in Kn) of some non-binary variable x[j] does not exist, then
1.663 +* formally b = +oo, in which case further analysis is not performed.
1.664 +*
1.665 +* Let Bp = {j in J: a[j] > 0}, Bn = {j in J: a[j] < 0}. To make all
1.666 +* the inequality coefficients in (5) positive, we replace all x[j] in
1.667 +* Bn by their complementaries, substituting x[j] = 1 - x~[j] for all
1.668 +* j in Bn, that gives:
1.669 +*
1.670 +* sum a[j] x[j] - sum a[j] x~[j] <= b~ - sum a[j]. (7)
1.671 +* j in Bp j in Bn j in Bn
1.672 +*
1.673 +* This inequality is a relaxation of the original constraint (1), and
1.674 +* it is a binary knapsack inequality. Writing it in the standard format
1.675 +* we have:
1.676 +*
1.677 +* sum alfa[j] z[j] <= beta, (8)
1.678 +* j in J
1.679 +*
1.680 +* where:
1.681 +* ( + a[j], if j in Bp,
1.682 +* alfa[j] = < (9)
1.683 +* ( - a[j], if j in Bn,
1.684 +*
1.685 +* ( x[j], if j in Bp,
1.686 +* z[j] = < (10)
1.687 +* ( 1 - x[j], if j in Bn,
1.688 +*
1.689 +* beta = b~ - sum a[j]. (11)
1.690 +* j in Bn
1.691 +*
1.692 +* In the inequality (8) all coefficients are positive, therefore, the
1.693 +* packing relaxation to be found for this inequality is the following:
1.694 +*
1.695 +* sum z[j] <= 1. (12)
1.696 +* j in P
1.697 +*
1.698 +* It is obvious that set P within J, which we would like to find, must
1.699 +* satisfy to the following condition:
1.700 +*
1.701 +* alfa[j] + alfa[k] > beta + eps for all j, k in P, j != k, (13)
1.702 +*
1.703 +* where eps is an absolute tolerance for value of the linear form.
1.704 +* Thus, it is natural to take P = {j: alpha[j] > (beta + eps) / 2}.
1.705 +* Moreover, if in the equality (8) there exist coefficients alfa[k],
1.706 +* for which alfa[k] <= (beta + eps) / 2, but which, nevertheless,
1.707 +* satisfies to the condition (13) for all j in P, *one* corresponding
1.708 +* variable z[k] (having, for example, maximal coefficient alfa[k]) can
1.709 +* be included in set P, that allows increasing the number of binary
1.710 +* variables in (12) by one.
1.711 +*
1.712 +* Once the set P has been built, for the inequality (12) we need to
1.713 +* perform back substitution according to (10) in order to express it
1.714 +* through the original binary variables. As the result of such back
1.715 +* substitution the relaxed packing inequality get its final format (2),
1.716 +* where Jp = J intersect Bp, and Jn = J intersect Bn. */
1.717 +
1.718 +int npp_implied_packing(NPP *npp, NPPROW *row, int which,
1.719 + NPPCOL *var[], char set[])
1.720 +{ struct elem *ptr, *e, *i, *k;
1.721 + int len = 0;
1.722 + double b, eps;
1.723 + /* build inequality (3) */
1.724 + if (which == 0)
1.725 + { ptr = copy_form(npp, row, -1.0);
1.726 + xassert(row->lb != -DBL_MAX);
1.727 + b = - row->lb;
1.728 + }
1.729 + else if (which == 1)
1.730 + { ptr = copy_form(npp, row, +1.0);
1.731 + xassert(row->ub != +DBL_MAX);
1.732 + b = + row->ub;
1.733 + }
1.734 + /* remove non-binary variables to build relaxed inequality (5);
1.735 + compute its right-hand side b~ with formula (6) */
1.736 + for (e = ptr; e != NULL; e = e->next)
1.737 + { if (!(e->xj->is_int && e->xj->lb == 0.0 && e->xj->ub == 1.0))
1.738 + { /* x[j] is non-binary variable */
1.739 + if (e->aj > 0.0)
1.740 + { if (e->xj->lb == -DBL_MAX) goto done;
1.741 + b -= e->aj * e->xj->lb;
1.742 + }
1.743 + else /* e->aj < 0.0 */
1.744 + { if (e->xj->ub == +DBL_MAX) goto done;
1.745 + b -= e->aj * e->xj->ub;
1.746 + }
1.747 + /* a[j] = 0 means that variable x[j] is removed */
1.748 + e->aj = 0.0;
1.749 + }
1.750 + }
1.751 + /* substitute x[j] = 1 - x~[j] to build knapsack inequality (8);
1.752 + compute its right-hand side beta with formula (11) */
1.753 + for (e = ptr; e != NULL; e = e->next)
1.754 + if (e->aj < 0.0) b -= e->aj;
1.755 + /* if beta is close to zero, the knapsack inequality is either
1.756 + infeasible or forcing inequality; this must never happen, so
1.757 + we skip further analysis */
1.758 + if (b < 1e-3) goto done;
1.759 + /* build set P as well as sets Jp and Jn, and determine x[k] as
1.760 + explained above in comments to the routine */
1.761 + eps = 1e-3 + 1e-6 * b;
1.762 + i = k = NULL;
1.763 + for (e = ptr; e != NULL; e = e->next)
1.764 + { /* note that alfa[j] = |a[j]| */
1.765 + if (fabs(e->aj) > 0.5 * (b + eps))
1.766 + { /* alfa[j] > (b + eps) / 2; include x[j] in set P, i.e. in
1.767 + set Jp or Jn */
1.768 + var[++len] = e->xj;
1.769 + set[len] = (char)(e->aj > 0.0 ? 0 : 1);
1.770 + /* alfa[i] = min alfa[j] over all j included in set P */
1.771 + if (i == NULL || fabs(i->aj) > fabs(e->aj)) i = e;
1.772 + }
1.773 + else if (fabs(e->aj) >= 1e-3)
1.774 + { /* alfa[k] = max alfa[j] over all j not included in set P;
1.775 + we skip coefficient a[j] if it is close to zero to avoid
1.776 + numerically unreliable results */
1.777 + if (k == NULL || fabs(k->aj) < fabs(e->aj)) k = e;
1.778 + }
1.779 + }
1.780 + /* if alfa[k] satisfies to condition (13) for all j in P, include
1.781 + x[k] in P */
1.782 + if (i != NULL && k != NULL && fabs(i->aj) + fabs(k->aj) > b + eps)
1.783 + { var[++len] = k->xj;
1.784 + set[len] = (char)(k->aj > 0.0 ? 0 : 1);
1.785 + }
1.786 + /* trivial packing inequality being redundant must never appear,
1.787 + so we just ignore it */
1.788 + if (len < 2) len = 0;
1.789 +done: drop_form(npp, ptr);
1.790 + return len;
1.791 +}
1.792 +
1.793 +/***********************************************************************
1.794 +* NAME
1.795 +*
1.796 +* npp_is_covering - test if constraint is covering inequality
1.797 +*
1.798 +* SYNOPSIS
1.799 +*
1.800 +* #include "glpnpp.h"
1.801 +* int npp_is_covering(NPP *npp, NPPROW *row);
1.802 +*
1.803 +* RETURNS
1.804 +*
1.805 +* If the specified row (constraint) is covering inequality (see below),
1.806 +* the routine npp_is_covering returns non-zero. Otherwise, it returns
1.807 +* zero.
1.808 +*
1.809 +* COVERING INEQUALITIES
1.810 +*
1.811 +* In canonical format the covering inequality is the following:
1.812 +*
1.813 +* sum x[j] >= 1, (1)
1.814 +* j in J
1.815 +*
1.816 +* where all variables x[j] are binary. This inequality expresses the
1.817 +* condition that in any integer feasible solution variables in set J
1.818 +* cannot be all equal to zero at the same time, i.e. at least one
1.819 +* variable must take non-zero (unity) value. W.l.o.g. it is assumed
1.820 +* that |J| >= 2, because if J is empty, the inequality (1) is
1.821 +* infeasible, and if |J| = 1, the inequality (1) is a forcing row.
1.822 +*
1.823 +* In general case the covering inequality may include original
1.824 +* variables x[j] as well as their complements x~[j]:
1.825 +*
1.826 +* sum x[j] + sum x~[j] >= 1, (2)
1.827 +* j in Jp j in Jn
1.828 +*
1.829 +* where Jp and Jn are not intersected. Therefore, using substitution
1.830 +* x~[j] = 1 - x[j] gives the packing inequality in generalized format:
1.831 +*
1.832 +* sum x[j] - sum x[j] >= 1 - |Jn|. (3)
1.833 +* j in Jp j in Jn
1.834 +*
1.835 +* (May note that the inequality (3) cuts off infeasible solutions,
1.836 +* where x[j] = 0 for all j in Jp and x[j] = 1 for all j in Jn.)
1.837 +*
1.838 +* NOTE: If |J| = 2, the inequality (3) is equivalent to packing
1.839 +* inequality (see the routine npp_is_packing). */
1.840 +
1.841 +int npp_is_covering(NPP *npp, NPPROW *row)
1.842 +{ /* test if constraint is covering inequality */
1.843 + NPPCOL *col;
1.844 + NPPAIJ *aij;
1.845 + int b;
1.846 + xassert(npp == npp);
1.847 + if (!(row->lb != -DBL_MAX && row->ub == +DBL_MAX))
1.848 + return 0;
1.849 + b = 1;
1.850 + for (aij = row->ptr; aij != NULL; aij = aij->r_next)
1.851 + { col = aij->col;
1.852 + if (!(col->is_int && col->lb == 0.0 && col->ub == 1.0))
1.853 + return 0;
1.854 + if (aij->val == +1.0)
1.855 + ;
1.856 + else if (aij->val == -1.0)
1.857 + b--;
1.858 + else
1.859 + return 0;
1.860 + }
1.861 + if (row->lb != (double)b) return 0;
1.862 + return 1;
1.863 +}
1.864 +
1.865 +/***********************************************************************
1.866 +* NAME
1.867 +*
1.868 +* npp_hidden_covering - identify hidden covering inequality
1.869 +*
1.870 +* SYNOPSIS
1.871 +*
1.872 +* #include "glpnpp.h"
1.873 +* int npp_hidden_covering(NPP *npp, NPPROW *row);
1.874 +*
1.875 +* DESCRIPTION
1.876 +*
1.877 +* The routine npp_hidden_covering processes specified inequality
1.878 +* constraint, which includes only binary variables, and the number of
1.879 +* the variables is not less than three. If the original inequality is
1.880 +* equivalent to a covering inequality (see below), the routine
1.881 +* replaces it by the equivalent inequality. If the original constraint
1.882 +* is double-sided inequality, it is replaced by a pair of single-sided
1.883 +* inequalities, if necessary.
1.884 +*
1.885 +* RETURNS
1.886 +*
1.887 +* If the original inequality constraint was replaced by equivalent
1.888 +* covering inequality, the routine npp_hidden_covering returns
1.889 +* non-zero. Otherwise, it returns zero.
1.890 +*
1.891 +* PROBLEM TRANSFORMATION
1.892 +*
1.893 +* Consider an inequality constraint:
1.894 +*
1.895 +* sum a[j] x[j] >= b, (1)
1.896 +* j in J
1.897 +*
1.898 +* where all variables x[j] are binary, and |J| >= 3. (In case of '<='
1.899 +* inequality it can be transformed to '>=' format by multiplying both
1.900 +* its sides by -1.)
1.901 +*
1.902 +* Let Jp = {j: a[j] > 0}, Jn = {j: a[j] < 0}. Performing substitution
1.903 +* x[j] = 1 - x~[j] for all j in Jn, we have:
1.904 +*
1.905 +* sum a[j] x[j] >= b ==>
1.906 +* j in J
1.907 +*
1.908 +* sum a[j] x[j] + sum a[j] x[j] >= b ==>
1.909 +* j in Jp j in Jn
1.910 +*
1.911 +* sum a[j] x[j] + sum a[j] (1 - x~[j]) >= b ==>
1.912 +* j in Jp j in Jn
1.913 +*
1.914 +* sum m a[j] x[j] - sum a[j] x~[j] >= b - sum a[j].
1.915 +* j in Jp j in Jn j in Jn
1.916 +*
1.917 +* Thus, meaning the transformation above, we can assume that in
1.918 +* inequality (1) all coefficients a[j] are positive. Moreover, we can
1.919 +* assume that b > 0, because otherwise the inequality (1) would be
1.920 +* redundant (see the routine npp_analyze_row). It is then obvious that
1.921 +* constraint (1) is equivalent to covering inequality only if:
1.922 +*
1.923 +* a[j] >= b, (2)
1.924 +*
1.925 +* for all j in J.
1.926 +*
1.927 +* Once the original inequality (1) is replaced by equivalent covering
1.928 +* inequality, we need to perform back substitution x~[j] = 1 - x[j] for
1.929 +* all j in Jn (see above).
1.930 +*
1.931 +* RECOVERING SOLUTION
1.932 +*
1.933 +* None needed. */
1.934 +
1.935 +static int hidden_covering(NPP *npp, struct elem *ptr, double *_b)
1.936 +{ /* process inequality constraint: sum a[j] x[j] >= b;
1.937 + 0 - specified row is NOT hidden covering inequality;
1.938 + 1 - specified row is covering inequality;
1.939 + 2 - specified row is hidden covering inequality. */
1.940 + struct elem *e;
1.941 + int neg;
1.942 + double b = *_b, eps;
1.943 + xassert(npp == npp);
1.944 + /* a[j] must be non-zero, x[j] must be binary, for all j in J */
1.945 + for (e = ptr; e != NULL; e = e->next)
1.946 + { xassert(e->aj != 0.0);
1.947 + xassert(e->xj->is_int);
1.948 + xassert(e->xj->lb == 0.0 && e->xj->ub == 1.0);
1.949 + }
1.950 + /* check if the specified inequality constraint already has the
1.951 + form of covering inequality */
1.952 + neg = 0; /* neg is |Jn| */
1.953 + for (e = ptr; e != NULL; e = e->next)
1.954 + { if (e->aj == +1.0)
1.955 + ;
1.956 + else if (e->aj == -1.0)
1.957 + neg++;
1.958 + else
1.959 + break;
1.960 + }
1.961 + if (e == NULL)
1.962 + { /* all coefficients a[j] are +1 or -1; check rhs b */
1.963 + if (b == (double)(1 - neg))
1.964 + { /* it is covering inequality; no processing is needed */
1.965 + return 1;
1.966 + }
1.967 + }
1.968 + /* substitute x[j] = 1 - x~[j] for all j in Jn to make all a[j]
1.969 + positive; the result is a~[j] = |a[j]| and new rhs b */
1.970 + for (e = ptr; e != NULL; e = e->next)
1.971 + if (e->aj < 0) b -= e->aj;
1.972 + /* now a[j] > 0 for all j in J (actually |a[j]| are used) */
1.973 + /* if b <= 0, skip processing--this case must not appear */
1.974 + if (b < 1e-3) return 0;
1.975 + /* now a[j] > 0 for all j in J, and b > 0 */
1.976 + /* the specified constraint is equivalent to covering inequality
1.977 + iff a[j] >= b for all j in J */
1.978 + eps = 1e-9 + 1e-12 * fabs(b);
1.979 + for (e = ptr; e != NULL; e = e->next)
1.980 + if (fabs(e->aj) < b - eps) return 0;
1.981 + /* perform back substitution x~[j] = 1 - x[j] and construct the
1.982 + final equivalent covering inequality in generalized format */
1.983 + b = 1.0;
1.984 + for (e = ptr; e != NULL; e = e->next)
1.985 + { if (e->aj > 0.0)
1.986 + e->aj = +1.0;
1.987 + else /* e->aj < 0.0 */
1.988 + e->aj = -1.0, b -= 1.0;
1.989 + }
1.990 + *_b = b;
1.991 + return 2;
1.992 +}
1.993 +
1.994 +int npp_hidden_covering(NPP *npp, NPPROW *row)
1.995 +{ /* identify hidden covering inequality */
1.996 + NPPROW *copy;
1.997 + NPPAIJ *aij;
1.998 + struct elem *ptr, *e;
1.999 + int kase, ret, count = 0;
1.1000 + double b;
1.1001 + /* the row must be inequality constraint */
1.1002 + xassert(row->lb < row->ub);
1.1003 + for (kase = 0; kase <= 1; kase++)
1.1004 + { if (kase == 0)
1.1005 + { /* process row lower bound */
1.1006 + if (row->lb == -DBL_MAX) continue;
1.1007 + ptr = copy_form(npp, row, +1.0);
1.1008 + b = + row->lb;
1.1009 + }
1.1010 + else
1.1011 + { /* process row upper bound */
1.1012 + if (row->ub == +DBL_MAX) continue;
1.1013 + ptr = copy_form(npp, row, -1.0);
1.1014 + b = - row->ub;
1.1015 + }
1.1016 + /* now the inequality has the form "sum a[j] x[j] >= b" */
1.1017 + ret = hidden_covering(npp, ptr, &b);
1.1018 + xassert(0 <= ret && ret <= 2);
1.1019 + if (kase == 1 && ret == 1 || ret == 2)
1.1020 + { /* the original inequality has been identified as hidden
1.1021 + covering inequality */
1.1022 + count++;
1.1023 +#ifdef GLP_DEBUG
1.1024 + xprintf("Original constraint:\n");
1.1025 + for (aij = row->ptr; aij != NULL; aij = aij->r_next)
1.1026 + xprintf(" %+g x%d", aij->val, aij->col->j);
1.1027 + if (row->lb != -DBL_MAX) xprintf(", >= %g", row->lb);
1.1028 + if (row->ub != +DBL_MAX) xprintf(", <= %g", row->ub);
1.1029 + xprintf("\n");
1.1030 + xprintf("Equivalent covering inequality:\n");
1.1031 + for (e = ptr; e != NULL; e = e->next)
1.1032 + xprintf(" %sx%d", e->aj > 0.0 ? "+" : "-", e->xj->j);
1.1033 + xprintf(", >= %g\n", b);
1.1034 +#endif
1.1035 + if (row->lb == -DBL_MAX || row->ub == +DBL_MAX)
1.1036 + { /* the original row is single-sided inequality; no copy
1.1037 + is needed */
1.1038 + copy = NULL;
1.1039 + }
1.1040 + else
1.1041 + { /* the original row is double-sided inequality; we need
1.1042 + to create its copy for other bound before replacing it
1.1043 + with the equivalent inequality */
1.1044 + copy = npp_add_row(npp);
1.1045 + if (kase == 0)
1.1046 + { /* the copy is for upper bound */
1.1047 + copy->lb = -DBL_MAX, copy->ub = row->ub;
1.1048 + }
1.1049 + else
1.1050 + { /* the copy is for lower bound */
1.1051 + copy->lb = row->lb, copy->ub = +DBL_MAX;
1.1052 + }
1.1053 + /* copy original row coefficients */
1.1054 + for (aij = row->ptr; aij != NULL; aij = aij->r_next)
1.1055 + npp_add_aij(npp, copy, aij->col, aij->val);
1.1056 + }
1.1057 + /* replace the original inequality by equivalent one */
1.1058 + npp_erase_row(npp, row);
1.1059 + row->lb = b, row->ub = +DBL_MAX;
1.1060 + for (e = ptr; e != NULL; e = e->next)
1.1061 + npp_add_aij(npp, row, e->xj, e->aj);
1.1062 + /* continue processing upper bound for the copy */
1.1063 + if (copy != NULL) row = copy;
1.1064 + }
1.1065 + drop_form(npp, ptr);
1.1066 + }
1.1067 + return count;
1.1068 +}
1.1069 +
1.1070 +/***********************************************************************
1.1071 +* NAME
1.1072 +*
1.1073 +* npp_is_partitioning - test if constraint is partitioning equality
1.1074 +*
1.1075 +* SYNOPSIS
1.1076 +*
1.1077 +* #include "glpnpp.h"
1.1078 +* int npp_is_partitioning(NPP *npp, NPPROW *row);
1.1079 +*
1.1080 +* RETURNS
1.1081 +*
1.1082 +* If the specified row (constraint) is partitioning equality (see
1.1083 +* below), the routine npp_is_partitioning returns non-zero. Otherwise,
1.1084 +* it returns zero.
1.1085 +*
1.1086 +* PARTITIONING EQUALITIES
1.1087 +*
1.1088 +* In canonical format the partitioning equality is the following:
1.1089 +*
1.1090 +* sum x[j] = 1, (1)
1.1091 +* j in J
1.1092 +*
1.1093 +* where all variables x[j] are binary. This equality expresses the
1.1094 +* condition that in any integer feasible solution exactly one variable
1.1095 +* in set J must take non-zero (unity) value while other variables must
1.1096 +* be equal to zero. W.l.o.g. it is assumed that |J| >= 2, because if
1.1097 +* J is empty, the inequality (1) is infeasible, and if |J| = 1, the
1.1098 +* inequality (1) is a fixing row.
1.1099 +*
1.1100 +* In general case the partitioning equality may include original
1.1101 +* variables x[j] as well as their complements x~[j]:
1.1102 +*
1.1103 +* sum x[j] + sum x~[j] = 1, (2)
1.1104 +* j in Jp j in Jn
1.1105 +*
1.1106 +* where Jp and Jn are not intersected. Therefore, using substitution
1.1107 +* x~[j] = 1 - x[j] leads to the partitioning equality in generalized
1.1108 +* format:
1.1109 +*
1.1110 +* sum x[j] - sum x[j] = 1 - |Jn|. (3)
1.1111 +* j in Jp j in Jn */
1.1112 +
1.1113 +int npp_is_partitioning(NPP *npp, NPPROW *row)
1.1114 +{ /* test if constraint is partitioning equality */
1.1115 + NPPCOL *col;
1.1116 + NPPAIJ *aij;
1.1117 + int b;
1.1118 + xassert(npp == npp);
1.1119 + if (row->lb != row->ub) return 0;
1.1120 + b = 1;
1.1121 + for (aij = row->ptr; aij != NULL; aij = aij->r_next)
1.1122 + { col = aij->col;
1.1123 + if (!(col->is_int && col->lb == 0.0 && col->ub == 1.0))
1.1124 + return 0;
1.1125 + if (aij->val == +1.0)
1.1126 + ;
1.1127 + else if (aij->val == -1.0)
1.1128 + b--;
1.1129 + else
1.1130 + return 0;
1.1131 + }
1.1132 + if (row->lb != (double)b) return 0;
1.1133 + return 1;
1.1134 +}
1.1135 +
1.1136 +/***********************************************************************
1.1137 +* NAME
1.1138 +*
1.1139 +* npp_reduce_ineq_coef - reduce inequality constraint coefficients
1.1140 +*
1.1141 +* SYNOPSIS
1.1142 +*
1.1143 +* #include "glpnpp.h"
1.1144 +* int npp_reduce_ineq_coef(NPP *npp, NPPROW *row);
1.1145 +*
1.1146 +* DESCRIPTION
1.1147 +*
1.1148 +* The routine npp_reduce_ineq_coef processes specified inequality
1.1149 +* constraint attempting to replace it by an equivalent constraint,
1.1150 +* where magnitude of coefficients at binary variables is smaller than
1.1151 +* in the original constraint. If the inequality is double-sided, it is
1.1152 +* replaced by a pair of single-sided inequalities, if necessary.
1.1153 +*
1.1154 +* RETURNS
1.1155 +*
1.1156 +* The routine npp_reduce_ineq_coef returns the number of coefficients
1.1157 +* reduced.
1.1158 +*
1.1159 +* BACKGROUND
1.1160 +*
1.1161 +* Consider an inequality constraint:
1.1162 +*
1.1163 +* sum a[j] x[j] >= b. (1)
1.1164 +* j in J
1.1165 +*
1.1166 +* (In case of '<=' inequality it can be transformed to '>=' format by
1.1167 +* multiplying both its sides by -1.) Let x[k] be a binary variable;
1.1168 +* other variables can be integer as well as continuous. We can write
1.1169 +* constraint (1) as follows:
1.1170 +*
1.1171 +* a[k] x[k] + t[k] >= b, (2)
1.1172 +*
1.1173 +* where:
1.1174 +*
1.1175 +* t[k] = sum a[j] x[j]. (3)
1.1176 +* j in J\{k}
1.1177 +*
1.1178 +* Since x[k] is binary, constraint (2) is equivalent to disjunction of
1.1179 +* the following two constraints:
1.1180 +*
1.1181 +* x[k] = 0, t[k] >= b (4)
1.1182 +*
1.1183 +* OR
1.1184 +*
1.1185 +* x[k] = 1, t[k] >= b - a[k]. (5)
1.1186 +*
1.1187 +* Let also that for the partial sum t[k] be known some its implied
1.1188 +* lower bound inf t[k].
1.1189 +*
1.1190 +* Case a[k] > 0. Let inf t[k] < b, since otherwise both constraints
1.1191 +* (4) and (5) and therefore constraint (2) are redundant.
1.1192 +* If inf t[k] > b - a[k], only constraint (5) is redundant, in which
1.1193 +* case it can be replaced with the following redundant and therefore
1.1194 +* equivalent constraint:
1.1195 +*
1.1196 +* t[k] >= b - a'[k] = inf t[k], (6)
1.1197 +*
1.1198 +* where:
1.1199 +*
1.1200 +* a'[k] = b - inf t[k]. (7)
1.1201 +*
1.1202 +* Thus, the original constraint (2) is equivalent to the following
1.1203 +* constraint with coefficient at variable x[k] changed:
1.1204 +*
1.1205 +* a'[k] x[k] + t[k] >= b. (8)
1.1206 +*
1.1207 +* From inf t[k] < b it follows that a'[k] > 0, i.e. the coefficient
1.1208 +* at x[k] keeps its sign. And from inf t[k] > b - a[k] it follows that
1.1209 +* a'[k] < a[k], i.e. the coefficient reduces in magnitude.
1.1210 +*
1.1211 +* Case a[k] < 0. Let inf t[k] < b - a[k], since otherwise both
1.1212 +* constraints (4) and (5) and therefore constraint (2) are redundant.
1.1213 +* If inf t[k] > b, only constraint (4) is redundant, in which case it
1.1214 +* can be replaced with the following redundant and therefore equivalent
1.1215 +* constraint:
1.1216 +*
1.1217 +* t[k] >= b' = inf t[k]. (9)
1.1218 +*
1.1219 +* Rewriting constraint (5) as follows:
1.1220 +*
1.1221 +* t[k] >= b - a[k] = b' - a'[k], (10)
1.1222 +*
1.1223 +* where:
1.1224 +*
1.1225 +* a'[k] = a[k] + b' - b = a[k] + inf t[k] - b, (11)
1.1226 +*
1.1227 +* we can see that disjunction of constraint (9) and (10) is equivalent
1.1228 +* to disjunction of constraint (4) and (5), from which it follows that
1.1229 +* the original constraint (2) is equivalent to the following constraint
1.1230 +* with both coefficient at variable x[k] and right-hand side changed:
1.1231 +*
1.1232 +* a'[k] x[k] + t[k] >= b'. (12)
1.1233 +*
1.1234 +* From inf t[k] < b - a[k] it follows that a'[k] < 0, i.e. the
1.1235 +* coefficient at x[k] keeps its sign. And from inf t[k] > b it follows
1.1236 +* that a'[k] > a[k], i.e. the coefficient reduces in magnitude.
1.1237 +*
1.1238 +* PROBLEM TRANSFORMATION
1.1239 +*
1.1240 +* In the routine npp_reduce_ineq_coef the following implied lower
1.1241 +* bound of the partial sum (3) is used:
1.1242 +*
1.1243 +* inf t[k] = sum a[j] l[j] + sum a[j] u[j], (13)
1.1244 +* j in Jp\{k} k in Jn\{k}
1.1245 +*
1.1246 +* where Jp = {j : a[j] > 0}, Jn = {j : a[j] < 0}, l[j] and u[j] are
1.1247 +* lower and upper bounds, resp., of variable x[j].
1.1248 +*
1.1249 +* In order to compute inf t[k] more efficiently, the following formula,
1.1250 +* which is equivalent to (13), is actually used:
1.1251 +*
1.1252 +* ( h - a[k] l[k] = h, if a[k] > 0,
1.1253 +* inf t[k] = < (14)
1.1254 +* ( h - a[k] u[k] = h - a[k], if a[k] < 0,
1.1255 +*
1.1256 +* where:
1.1257 +*
1.1258 +* h = sum a[j] l[j] + sum a[j] u[j] (15)
1.1259 +* j in Jp j in Jn
1.1260 +*
1.1261 +* is the implied lower bound of row (1).
1.1262 +*
1.1263 +* Reduction of positive coefficient (a[k] > 0) does not change value
1.1264 +* of h, since l[k] = 0. In case of reduction of negative coefficient
1.1265 +* (a[k] < 0) from (11) it follows that:
1.1266 +*
1.1267 +* delta a[k] = a'[k] - a[k] = inf t[k] - b (> 0), (16)
1.1268 +*
1.1269 +* so new value of h (accounting that u[k] = 1) can be computed as
1.1270 +* follows:
1.1271 +*
1.1272 +* h := h + delta a[k] = h + (inf t[k] - b). (17)
1.1273 +*
1.1274 +* RECOVERING SOLUTION
1.1275 +*
1.1276 +* None needed. */
1.1277 +
1.1278 +static int reduce_ineq_coef(NPP *npp, struct elem *ptr, double *_b)
1.1279 +{ /* process inequality constraint: sum a[j] x[j] >= b */
1.1280 + /* returns: the number of coefficients reduced */
1.1281 + struct elem *e;
1.1282 + int count = 0;
1.1283 + double h, inf_t, new_a, b = *_b;
1.1284 + xassert(npp == npp);
1.1285 + /* compute h; see (15) */
1.1286 + h = 0.0;
1.1287 + for (e = ptr; e != NULL; e = e->next)
1.1288 + { if (e->aj > 0.0)
1.1289 + { if (e->xj->lb == -DBL_MAX) goto done;
1.1290 + h += e->aj * e->xj->lb;
1.1291 + }
1.1292 + else /* e->aj < 0.0 */
1.1293 + { if (e->xj->ub == +DBL_MAX) goto done;
1.1294 + h += e->aj * e->xj->ub;
1.1295 + }
1.1296 + }
1.1297 + /* perform reduction of coefficients at binary variables */
1.1298 + for (e = ptr; e != NULL; e = e->next)
1.1299 + { /* skip non-binary variable */
1.1300 + if (!(e->xj->is_int && e->xj->lb == 0.0 && e->xj->ub == 1.0))
1.1301 + continue;
1.1302 + if (e->aj > 0.0)
1.1303 + { /* compute inf t[k]; see (14) */
1.1304 + inf_t = h;
1.1305 + if (b - e->aj < inf_t && inf_t < b)
1.1306 + { /* compute reduced coefficient a'[k]; see (7) */
1.1307 + new_a = b - inf_t;
1.1308 + if (new_a >= +1e-3 &&
1.1309 + e->aj - new_a >= 0.01 * (1.0 + e->aj))
1.1310 + { /* accept a'[k] */
1.1311 +#ifdef GLP_DEBUG
1.1312 + xprintf("+");
1.1313 +#endif
1.1314 + e->aj = new_a;
1.1315 + count++;
1.1316 + }
1.1317 + }
1.1318 + }
1.1319 + else /* e->aj < 0.0 */
1.1320 + { /* compute inf t[k]; see (14) */
1.1321 + inf_t = h - e->aj;
1.1322 + if (b < inf_t && inf_t < b - e->aj)
1.1323 + { /* compute reduced coefficient a'[k]; see (11) */
1.1324 + new_a = e->aj + (inf_t - b);
1.1325 + if (new_a <= -1e-3 &&
1.1326 + new_a - e->aj >= 0.01 * (1.0 - e->aj))
1.1327 + { /* accept a'[k] */
1.1328 +#ifdef GLP_DEBUG
1.1329 + xprintf("-");
1.1330 +#endif
1.1331 + e->aj = new_a;
1.1332 + /* update h; see (17) */
1.1333 + h += (inf_t - b);
1.1334 + /* compute b'; see (9) */
1.1335 + b = inf_t;
1.1336 + count++;
1.1337 + }
1.1338 + }
1.1339 + }
1.1340 + }
1.1341 + *_b = b;
1.1342 +done: return count;
1.1343 +}
1.1344 +
1.1345 +int npp_reduce_ineq_coef(NPP *npp, NPPROW *row)
1.1346 +{ /* reduce inequality constraint coefficients */
1.1347 + NPPROW *copy;
1.1348 + NPPAIJ *aij;
1.1349 + struct elem *ptr, *e;
1.1350 + int kase, count[2];
1.1351 + double b;
1.1352 + /* the row must be inequality constraint */
1.1353 + xassert(row->lb < row->ub);
1.1354 + count[0] = count[1] = 0;
1.1355 + for (kase = 0; kase <= 1; kase++)
1.1356 + { if (kase == 0)
1.1357 + { /* process row lower bound */
1.1358 + if (row->lb == -DBL_MAX) continue;
1.1359 +#ifdef GLP_DEBUG
1.1360 + xprintf("L");
1.1361 +#endif
1.1362 + ptr = copy_form(npp, row, +1.0);
1.1363 + b = + row->lb;
1.1364 + }
1.1365 + else
1.1366 + { /* process row upper bound */
1.1367 + if (row->ub == +DBL_MAX) continue;
1.1368 +#ifdef GLP_DEBUG
1.1369 + xprintf("U");
1.1370 +#endif
1.1371 + ptr = copy_form(npp, row, -1.0);
1.1372 + b = - row->ub;
1.1373 + }
1.1374 + /* now the inequality has the form "sum a[j] x[j] >= b" */
1.1375 + count[kase] = reduce_ineq_coef(npp, ptr, &b);
1.1376 + if (count[kase] > 0)
1.1377 + { /* the original inequality has been replaced by equivalent
1.1378 + one with coefficients reduced */
1.1379 + if (row->lb == -DBL_MAX || row->ub == +DBL_MAX)
1.1380 + { /* the original row is single-sided inequality; no copy
1.1381 + is needed */
1.1382 + copy = NULL;
1.1383 + }
1.1384 + else
1.1385 + { /* the original row is double-sided inequality; we need
1.1386 + to create its copy for other bound before replacing it
1.1387 + with the equivalent inequality */
1.1388 +#ifdef GLP_DEBUG
1.1389 + xprintf("*");
1.1390 +#endif
1.1391 + copy = npp_add_row(npp);
1.1392 + if (kase == 0)
1.1393 + { /* the copy is for upper bound */
1.1394 + copy->lb = -DBL_MAX, copy->ub = row->ub;
1.1395 + }
1.1396 + else
1.1397 + { /* the copy is for lower bound */
1.1398 + copy->lb = row->lb, copy->ub = +DBL_MAX;
1.1399 + }
1.1400 + /* copy original row coefficients */
1.1401 + for (aij = row->ptr; aij != NULL; aij = aij->r_next)
1.1402 + npp_add_aij(npp, copy, aij->col, aij->val);
1.1403 + }
1.1404 + /* replace the original inequality by equivalent one */
1.1405 + npp_erase_row(npp, row);
1.1406 + row->lb = b, row->ub = +DBL_MAX;
1.1407 + for (e = ptr; e != NULL; e = e->next)
1.1408 + npp_add_aij(npp, row, e->xj, e->aj);
1.1409 + /* continue processing upper bound for the copy */
1.1410 + if (copy != NULL) row = copy;
1.1411 + }
1.1412 + drop_form(npp, ptr);
1.1413 + }
1.1414 + return count[0] + count[1];
1.1415 +}
1.1416 +
1.1417 +/* eof */